The ionization enthalpy of `Na^+` formation from Nad is `495.8 kJ mol^(-1)`, while the electron gain enthalpy of Bris `-325.0 kJ mol^(-1)` .Given the lattice enthalpy of NaBr is `-728.4 kJ mol^(-1)` The energy for the formation of NaBr ionic solid is `(-)"______" xx 10^(-1)kJ "mol"^(-1)`

Answer

To find the energy for the formation of NaBr ionic solid, we will use the following equation based on the Born-Haber cycle:

\[ \Delta H_f^\circ = \Delta H_{ionization} + \Delta H_{electron \, gain} + \Delta H_{lattice} \]

Where:
- \(\Delta H_f^\circ\) is the enthalpy of formation of NaBr.
- \(\Delta H_{ionization}\) is the ionization enthalpy of Na.
- \(\Delta H_{electron \, gain}\) is the electron gain enthalpy of Br.
- \(\Delta H_{lattice}\) is the lattice enthalpy of NaBr.

### Step 1: Write down the given values
- Ionization enthalpy of Na: \(\Delta H_{ionization} = 495.8 \, kJ \, mol^{-1}\)
- Electron gain enthalpy of Br: \(\Delta H_{electron \, gain} = -325.0 \, kJ \, mol^{-1}\)
- Lattice enthalpy of NaBr: \(\Delta H_{lattice} = -728.4 \, kJ \, mol^{-1}\)

### Step 2: Substitute the values into the equation
Now we substitute the values into the Born-Haber cycle equation:

\[ \Delta H_f^\circ = 495.8 + (-325.0) + (-728.4) \]

### Step 3: Perform the calculations
1. Calculate \(495.8 - 325.0\):
\[ 495.8 - 325.0 = 170.8 \]

2. Now add the lattice enthalpy:
\[ 170.8 - 728.4 = -557.6 \]

### Step 4: Final result
Thus, the energy for the formation of NaBr ionic solid is:
\[ \Delta H_f^\circ = -557.6 \, kJ \, mol^{-1} \]

To express it in the required format:
\[ \Delta H_f^\circ = -5.576 \times 10^2 \, kJ \, mol^{-1} \]

### Answer:
The energy for the formation of NaBr ionic solid is \(-5.576 \times 10^2 \, kJ \, mol^{-1}\).